= Solution
Use the metric $g^{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$. Transposing a product reverses its order but does not conjugate the factor $i$. Thus the <charge-conjugation matrix> identity gives
$$
C\gamma_5^TC^{-1}=i(-\gamma^3)(-\gamma^2)(-\gamma^1)(-\gamma^0)=i\gamma^0\gamma^1\gamma^2\gamma^3=\boxed{\gamma_5},
$$
where reversing the four distinct <gamma matrices> requires six exchanges. Similarly,
$$
C[\gamma^\mu,\gamma^\nu]^TC^{-1}=\gamma^\nu\gamma^\mu-\gamma^\mu\gamma^\nu=\boxed{-[\gamma^\mu,\gamma^\nu]}.
$$
To fix the transpose sign, work in the <Weyl representation of the gamma matrices>:
$$
\gamma^0=\begin{pmatrix}0&I\\I&0\end{pmatrix},\qquad \gamma^j=\begin{pmatrix}0&\tau_j\\-\tau_j&0\end{pmatrix},\qquad C_0=i\gamma^2\gamma^0=\begin{pmatrix}\varepsilon&0\\0&-\varepsilon\end{pmatrix},\quad\varepsilon=i\tau_2.
$$
The Pauli identity checks $C_0\gamma^{\mu T}C_0^{-1}=-\gamma^\mu$ directly, and $\varepsilon^T=-\varepsilon$ gives $C_0^T=-C_0$, with $C_0^\dagger C_0=I$. Any other intertwiner differs from $C_0$ by a scalar: its ratio with $C_0$ commutes with all <gamma matrices>, whose products span the full four-by-four matrix algebra. A matrix commuting with all matrix units is scalar. Under a spinor basis change $S$, the intertwiner becomes $SC_0S^T$, whose transpose is still its negative. Therefore in every such four-component representation
$$
\boxed{C^T=-C.}
$$
This proves the <antisymmetry of the four-dimensional charge-conjugation matrix>, rather than leaving the sign undetermined by the general transpose argument.
Hermitian conjugating the charged <Dirac equation>, using real $A_\mu,e,m$ and $\gamma^0\gamma^{\mu\dagger}\gamma^0=\gamma^\mu$, gives the <adjoint Dirac equation>
$$
-i(\partial_\mu\bar\psi)\gamma^\mu+eA_\mu\bar\psi\gamma^\mu-m\bar\psi=0.
$$
Transpose it and multiply by $C$. Since $C\gamma^{\mu T}=-\gamma^\mu C$, the field $\psi^c=C\bar\psi^T$ obtained by <charge conjugation> obeys
$$
\boxed{[i\gamma^\mu(\partial_\mu+ieA_\mu)-m]\psi^c=0.}
$$
Thus algebraic <charge conjugation> reverses the electromagnetic coupling.
If $\gamma_5\psi=\eta\psi$, $\eta=\pm1$, Hermiticity of $\gamma_5$ and its anticommutation with $\gamma^0$ imply
$$
\bar\psi\gamma_5=\psi^\dagger\gamma^0\gamma_5=-\psi^\dagger\gamma_5\gamma^0=-\eta\bar\psi.
$$
Using the first transpose identity,
$$
\gamma_5\psi^c=C\gamma_5^T\bar\psi^T=C(\bar\psi\gamma_5)^T=-\eta\psi^c.
$$
Hence \b[$\boxed{\bar\psi\gamma_5=-\eta\bar\psi,\quad\gamma_5\psi^c=-\eta\psi^c}$]: <algebraic charge conjugation reverses chirality>. This concerns conjugating the already projected spinor, not the distinct operation of keeping a chiral projector outside an operator conjugation.
For the <Grassmann variation of a chiral Majorana mass term>, variations of odd fields must be moved past odd fields with a minus sign. If $A^T=-A$, componentwise reordering gives
$$
\delta(\psi^TA\psi)=\delta\psi^TA\psi+\psi^TA\delta\psi=2\delta\psi^TA\psi.
$$
Both $C$ and $C^{-1}$ are antisymmetric. Treating $\psi,\bar\psi$ independently, integrating the kinetic variation by parts, and placing varied Grassmann fields on the left gives
$$
\delta\mathcal L=\delta\bar\psi(i\gamma^\mu\partial_\mu\psi-m^*C\bar\psi^T)+\delta\psi^T(i\gamma^{\mu T}\partial_\mu\bar\psi^T+mC^{-1}\psi)+\partial_\mu(\cdots).
$$
The factors two from variation cancel the mass-term factors $1/2$. The first coefficient gives one equation; multiplying the second by $C$ gives the other:
$$
\boxed{i\gamma^\mu\partial_\mu\psi-m^*\psi^c=0,\qquad i\gamma^\mu\partial_\mu\psi^c-m\psi=0.}
$$
The variations can be restricted to their chiral subspaces: the two terms in each equation have the same opposite output <chirality>, so no inactive component equation has been imposed.
Applying $i\not\partial$ to the first equation and substituting the second, using $(i\not\partial)^2=-\Box$, gives
$$
(\Box+|m|^2)\psi=0,\qquad\boxed{M=|m|.}
$$
A phase rotation removes the phase of $m$, and combining the chiral field with its conjugate then gives a <Majorana fermion> of this mass, not two independent chiral species.
Let $\chi=\phi'^\dagger\Psi$ and $\bar\chi=\bar\Psi\phi'$. Question 1 shows they are gauge singlets, so both terms in $\mathcal L_G=G\chi^TC^{-1}\chi-G^*\bar\chi C\bar\chi^T$ preserve <electroweak gauge invariance>. The original PDF has $C^{-1}$ in the first term; the TeX aid incorrectly replaces it by $G^{-1}$. At the <Higgs field> vacuum,
$$
\phi'_0=\frac v{\sqrt2}\begin{pmatrix}1\\0\end{pmatrix},\qquad \chi\longrightarrow\frac v{\sqrt2}\nu_L,
$$
so
$$
\mathcal L_{G,0}=\frac12\left[Gv^2\nu_L^TC^{-1}\nu_L-G^*v^2\bar\nu_LC\bar\nu_L^T\right].
$$
Comparing with the just-derived chiral mass equations yields the <Majorana neutrino mass from the Weinberg operator>:
$$
\boxed{m_{\rm parameter}=Gv^2,\qquad m_\nu=|G|v^2.}
$$
For nonzero $G$ this generates a mass without introducing a <right-handed neutrino>. Under <lepton number> rotation $\Psi\mapsto e^{i\alpha}\Psi$, the first operator acquires $e^{2i\alpha}$ and its conjugate acquires $e^{-2i\alpha}$. Thus \b[the interaction violates lepton number by $\boxed{\Delta L=2}$]. Each composite contains one scalar of mass dimension one and one fermion of dimension $3/2$, so the interaction is dimension five and $G$ has dimension $-1$. This is the <Weinberg operator>, an effective extension of the minimal renormalizable <Standard Model>, consistent with the absence of bare quadratic masses in Question 1.
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